from the newtons equation of motion :
"s=ut + \\dfrac{1}{2}at^{2}"
considering vertical projection initial velocity is = 0
"u= 0"
"200=0(t )+ \\dfrac{1}{2}(9.8)t^{2}"
"200 \\times 2= 0 + 9.8\\times t^2"
"t= \\sqrt \\dfrac{400}{9.8}= 6.389" seconds
estimated distance = horizontal velocity x time taken to hit the ground
"s= 25 m\/s \\times 6.389 seconds = 159.725 m"
for vertical motion the initial velocity =0 , assuming acceleration is 9.8 m/s
final velocity velocity with which the shell hits the target and the direction of shell at the time of hitting the target is obtained by the formula
"v=u +at""v= 0 + 9.8 \\times 6.389= 62.61 m\/s"
horizontal velocity will be kept constant at 25 m/s
but there will be downward acceleration of the shell to attain a maximum velocity of 62. 61 m/s at the time of hitting the target.
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